An algebra problem by Journel Cabrillos

Let the original price of the item be $x$ .

The item was already discounted by $10\%$ , which means the price was $\dfrac{90}{100} \times x = \dfrac{9}{10}x$

After that, the item was discounted by another $10\%$ , which causes the price to become $\dfrac{90}{100} \times \dfrac{9}{10}x = \dfrac{81}{100}x$

Calculate the total discount:

$\dfrac{x - \frac{81}{100}x}{x} \times 100\%\\ =\dfrac{\left(\frac{19}{100}x \right)}{x} \times 100\%\\ =\dfrac{19}{100} \times 100\%\\ =\boxed{19\%}$